Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 5

int 1/(x^2-9)dx
To solve using the partial fraction method, the first step is to factor the denominator of the integrand.
1/(x^2-9) =1/((x - 3)(x +3))
Then, express it as a sum of two fractions.
1/((x-3)(x+3))=A/(x-3)+B/(x+3)
To solve for the values of A and B, multiply both sides by the LCD.
(x-3)(x+3)*1/((x-3)(x+3))=(A/(x-3)+B/(x+3))*(x-3)(x+3)
1 = A(x+3)+B(x-3)
Then, assign values to x in such a way that either (x+3) or (x-3) will be zero. So, plug-in x = 3 to get the value of A.
1=A(3+3) + B(3-3)
1=A*6+B*0
1=6A
1/6=A
Also, plug-in x=-3 to get the value of B.
1=A(-3+3)+B(-3-3)
1=A*0 + B*(-6)
1=-6B
-1/6=B
So the partial fraction decomposition of the integrand is:
int 1/(x^2-9)dx=int (1/(6(x-3)) -1/(6(x+3)))dx
Then, express it as difference of two integrals.
=int 1/(6(x-3))dx - int 1/(6(x+3))dx
=1/6 int 1/(x-3)dx - 1/6 int 1/(x+3)dx
And, apply the integral formula int 1/u du = ln|u|+C .
=1/6ln|x-3| -1/6ln|x+3|+C

Therefore, int 1/(x^2-9)dx=1/6ln|x-3| -1/6ln|x+3|+C .